Question

CH 9 ITEM 11

A frictionless pulley has the shape of a uniform solid disk of mass 2.90 kg and radius 10 cm. A 1.70 kg stone is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest.

Part A

How far must the stone fall so that the pulley has 3.00 J of kinetic energy?

Part B

What percent of the total kinetic energy does the pulley have?

Answer 1

here,

mass of pulley , m = 2.9 kg

radius , r = 10 cm

r = 0.1 m

moment of inertia , I = 0.5 * m * r^2

I = 0.5 * 2.9 * 0.1^2

I = 0.0145 kg . m^2

mass of stone , M = 1.7 kg

part A)

let the angular velocity of pulley be w

v = r * w

KE = 3 J

0.5 * I * w^2 = 3

0.5 * 0.0145 * w^2 = 3

w = 20.34 rad/s

let trhe stone fall x distance

kinetic energy of pulley , KE = 3 J

using conservation of energy ,

kinetic energy of pulley + kinetic energy of stone = change in potential energy of stone

0.5 * M * v^2 + 0.5 * I * w^2 = M * g * h

0.5 * 1.7 * (0.1 * 20.34)^2 + 0.5 * 0.0145 * 20.34^2 = 1.7 * 9.8 * h

h = 0.39 m

the stone moves 0.39 m downward

part B)

kinetic energy of pulley ,KEp = 3 J

total kinetic energy , KEt = change in potential energy of stone

KEt = M * g * h

KEt = 1.7 * 9.8 * 0.39

KEt = 6.5 J

percentage of the total kinetic energy does the pulley have, % = KEp/KEt * 100

% = 3/6.5 * 100

% = 46.15 %

percentage of the total kinetic energy does the pulley have 46.15 %